# proof of some statement about finite field extension

I would like to have some explanation for the following statement

Let $$K$$ be an algebraically closed field of characteristic $$p>0$$, and $$K((t))$$, the field of Laurent series with coefficients in $$K$$. The Galois group of the polynomial $$X^{p^n}-X=t^{-1}$$ is isomorphic to the additive group of $$F_{p^n}$$, i.e. to $$(\mathbb{Z}/p\mathbb{Z})^n$$.

Another question:

Are there some extensions with Galois group isomorphic to $$(\mathbb{Z}/p^n\mathbb{Z})$$ with $$n>1$$

• It is very rude to just change the problem statement after the problem has been answered. – Dedalus Aug 4 at 9:49
• @Dedalus I'm sorry i wanted to post a new question, but got confused with this.. – Josh fisher Aug 4 at 9:52
• No worries, just wanted to inform you. – Dedalus Aug 4 at 9:54

Say that $$\alpha$$ is a root of your polynomial $$X^{p^n}-X-t^{-1}=0.$$ Then it is obvious that if $$a \in \mathbb{F}_{p^n},$$ that $$\alpha+a$$ is a root as well, since $$(\alpha+a)^{p^n}= \alpha^{p^n}+a^{p^n} = \alpha^{p^n}+a.$$ So the Galois group is as claimed.
There are finite extensions with Galois groups isomorphic to $$\mathbb{Z}/p^n\mathbb{Z}$$ with $$n>1.$$ This can be done using the Witt polynomials, see for example: Cyclic Artin-Schreier-Witt extension of order $p^2$ .
• Everything you say is, of course, correct. But the answer is missing the piece that the given polynomial is irreducible. That follows from Artin-Schreier theory if we can show that the polynomial has no roots in $K((t))$. Mind you, I'm sure you know how to handle that, but I'm not sure all the readers are aware of all the subtleties. For example, if we replace $t^{-1}$ with $t$, then the series $u=-(t+t^{p^n}+t^{2p^n}+\cdots)\in K((t))$ is a zero of $X^{p^n}-X-t$. – Jyrki Lahtonen Aug 4 at 10:19
• @Dedalus Do you mean to use the fact that $v_{\mathfrak{P}}(X^{p^n} - X) = v_{\mathfrak{P}}(T^{-1}) = e(\mathfrak{P}/T)$ ? – Josh fisher Aug 10 at 18:20